jmc

Substituting $f^{-1}(x)$ into our expression for $f$, we get $$\frac{3}{2-f^{-1}(x)}=x.$$Solving for $f^{-1}(x)$, we find that $f^{-1}(x)=2-\frac{3}{x}$, so $f^{-1}(3)=2-\frac{3}{3}=1$. Therefore, $g(3)=\frac{1}{f^{-1}(3)}+9=\frac{1}{1}+9=\boxed{10}$.